Show that any $n+1$ vectors in a $n$ dimensional vector space forms a linearly dependent set

$V$ is an $n$-dimensional vector space. Show that $n + 1$ vectors in $V$ form a linearly dependent set.

Here is how I am approaching it:

Let $\dim V = n$, which implies that $S$ is a linearly independent set of vectors such that $S = \{v_1,v_2,\ldots,v_n\}$ is the basis of $V$.

Let $W = \{w_1,\ldots,w_r\}$ be a set of linearly independent vectors in $V$

I don't know where to go from here.

• You need to use a basis for $V$. $V$ is not a set with $n$ vectors. It has infinitely many vectors. But if its dimension is $n$ then you can pick up a basis with $n$ LI vectors. Nov 2, 2013 at 16:32
• @Sigur Updated it to what I think your comment was leading me towards. Do you have any advice on where to go from here? Nov 2, 2013 at 16:40
• Can you assume that n of them are base, then imply that the (n+1)th vector can be displayed as linear combination of the others? Nov 2, 2013 at 16:45
• @LeeNeverGup That's what the question requires as an answer. I don't know how to show that. Nov 2, 2013 at 16:48
• It's better to use TeX syntax to type math.
– leo
Nov 2, 2013 at 18:04

The statement $\dim V = n$ means there exists a basis $\{v_1,\ldots,v_n\}$ for $V$. Let $v_{n+1} \in V\setminus \{v_1,\ldots,v_n\}$. Then by the definition of a basis, there exist $\alpha_1,\ldots,\alpha_n \in F$ ($F$ being the field over which $V$ is defined) such that $$v_{n+1} = \alpha_1v_1 + \ldots + \alpha_nv_n.$$ So $\{v_1,\ldots,v_{n+1}\}$ is linearly dependent.

• This doesn't prove "any" $n+1$ vectors is LD. Oct 25, 2014 at 23:44
• @LTS: As I recall, my intent with the answer was to let the OP reach the desired conclusion based on the above, since it follows almost immediately. Indeed, since we are assuming $V$ is of dimension $n$, every set of $n$ linearly independent vectors in $V$ is a basis for $V$. If $S \subset V$ is of order $n+1$, then either every $n$ element subset of $S$ is linearly dependent (in which case so is $S$) or $S$ contains a basis for $V$. The set of vectors is linearly dependent in the former case immediately and in the latter case based on my answer.
– Dan
Oct 26, 2014 at 19:53

Believe I can answer my own question:

Assuming the above let Let $W_1$ = {$w_1$, $v_1$, $v_2$, . . . $v_n$}

$W_1$ spans V because S spans V.

$w_1$ is a linear combination of the vectors in S because S is a basis for V.

Then we get $w_1$ = $a_1$$v_1 + a_2$$v_2$ + . . . + $a_n$$v_n rearrange terms so that 0 = a_1$$v_1$ + $a_2$$v_2 + . . . + a_n$$v_n$ - $w_1$ and $W_1$ is a linearly dependent set because at least one of the coefficients does not equal 0/